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2 years ago

Three identical lights are connected in series to a 12V battery. How does the brightness of each light compare?

Physics

1 answer:

kodGreya [7K]2 years ago

8 0

Answer: They are identical brightness

Explanation:

If the lights are assumed to be resistance bulbs

Each light has the same current and will each drop one third of the supply voltage.

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Is the energy of a wave is calculated by squaring the frequency of the wave?

Misha Larkins [42]

Nope.
Energy is directly proportional to frequency. and when you calculate energy, you multiply frequency with a constant number called "Planck's Constant"

E = hf

Hope this helps!

5 0

2 years ago

How do you increase the rate of dissolving ?

Bogdan [553]

To increase the rate of dissolving you can stir the concentration, heat the mixture, or grind the solute.

6 0

3 years ago

Read 2 more answers

All objects emit ______ radiation?<br> A-electromagnetic<br> B-kinetic<br> D-solar

natulia [17]

A, electromagnetic radiation

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3 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with

padilas [110]

Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s

A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s

B) momentum of the raft = - momentum of the girl = -24.0 N/s

C) speed of the raft

p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]

8 0

3 years ago